Can someone me? again! and you have to use pi(3.14)

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madeleine1102

09.11.2021

Not sure what you mean by "the answer is: ", which is definitely each successive term differs by 2. if we assign $k=1$ to the first term, then we can write $a_1=6$ $a_2=8=a_1+2$ $a_3=10=a_2+2=a_1+2\cdot2$ and so on, with the general $n$ -th term given by $a_n=a_1+2(n-1)$ the sum of the first 11 terms of this arithmetic sequence is $s_{11}=\displaystyle\sum_{k=1}^{11}(6+2(n-1))=\sum_{k=1}^{11}(2n+4)$ if you're supposed to find the value of the sum, then we have $\displaystyle\sum_{k=1}^nk=\frac{n(n+1)}2$ $\displaystyle\sum_{k=1}^n1=n$ from which we get $s_{11}=\displaystyle2\sum_{k=1}^{11}n+4\sum_{k=1}^{11}1$ $s_{11}=2\cdot\dfrac{11\cdot12}2+4\cdot11=176$

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Can someone me? again! and you have to use pi(3.14)

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